The K_eq for the equilibrium below in 0.112 at 700.0 degree C SO_2(g) + 1/2 O_2(

Question

The K_eq for the equilibrium below in 0.112 at 700.0 degree C SO_2(g) + 1/2 O_2(g) rlhar SO_3(g) What is the value of at this temperature for the following reaction? 2SO_3(g) rlhar 2SO_2(g) + O_2(g) 4 46 79.7 8.93 17.86 Which one of the following will change the value of an equilibrium constant? adding other substances that do not react with any of the species involved in the equilibrium varying the initial concentrations of products varying the initial concentrations of reactants changing temperature changing the volume of the reaction vessel Which one of the following is not a valid expression for the rate of the reaction below? 4NH_3 + 7O_2 rightarrow 4NO_2 + 6H_2O -1/7 delta[O_2]/delta t -1/4 [NH_3]/delta t 1/4 delta[NO_2]/delta t 1/6 delta[H_2O]/delta t All of the above am valid expressions of the reaction rate The equilibrium expression for Kp for the reaction below is ________. 2O_3(g) rlhar 3O_2(g) PO_2^3/PO^2_3 2PO_3/3PO_2 PO_3^2/PO_2^2 3PO_3/2PO_2

Explanation / Answer

14) as the reaction is reversed and multiplied with 2 , so new Keq= 1/(Keq)^2

=1/(0.112)^2 =79.72 option (D)

15) Keq depends on only temparature, option (D)

16) all are valid expressions hence option (E)

17) Keq = Po2 ^3 /Po3^2 , hence option is (A)

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